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Brooklyn 99 And The 12 Men On An Island Puzzle

I was watching Brooklyn Nine-Nine and the ‘Captain Peralta’ episode when they posed this problem:

Twelve men are on a desert island, all with identical weights except for one of them, who is slightly lighter or heavier than the others. The only other thing on the island is a seesaw. There are no scales or means to measure weight otherwise. Can you determine which man has the different weight? You only get to use the seesaw three times.

This is a version of the balance or twelve coin problem which is as old as the hills. Variations of it abound, and keeping them straight is a pain (and made for a few annoying minutes for me last night). Hence, here’s some solutions:

By the way, we’ll talk about coins numbered 01 through 12 from here on in – men, weights, marbles, it doesn’t matter. What is key is that you have a limited number of weighings, and only one is not like the others (optionally, you may know if it is lighter or heavier, but you may not).

Problem One: One Item is Heavier/Lighter. Find It.

Pretty easy if you know the coin is definitely heavier or lighter than the others, because the tilt of the scales gives you extra information (a lighter one will tip the scale up on that side, for example). So weigh 6 on one side, 6 on the other. Pick the lighter/heavier side which will contain your ‘bad’ coin. Then split those 6 into two groups of three, and do it again. The final three are tested this way: weigh 2 of them, one on each side. If it tilts, you know it’s the light/heavy one on the scale. If they balance, the odd one of the trio you didn’t weigh is the bad one. This final part is the ‘trick’, since some people are left wondering for the third step how you can weigh three items with one measurement – but as you can see, a balance between the two on the scale points the finger at the third one not on the scale…

This is the classic form of the puzzle, and likely the one referred to by Amy Santiago when she said “You place six on each side…”

Now it’s trickier – you don’t have the knowledge if it’s heavier or lighter, just that it is different. It can be done however. To make it easy, we’ll consider 12 coins, with one that is lighter or heavier, and numbered from 01 to 12, along with a pan scale:

Weighing 1: Weigh 4 on one side, 4 on the other. There’s three possible outcomes we care about here: The scale shows that the left side is heavier (LH), the right side is heavier (RH), or they are equal (EQ). So for01,02,03,04 versus 05,06,07,08 and ignoring 09,10,11,12
Let’s look at each in detail:

Equal (EQ): If these are equal, then they are all good coins, and one of 09,10,11,12 is the problem – but we don’t know if they’re heavier or lighter!
Weighing 2: Measure three of these coins against three of any of the others (which we now know are all good, so we’ll ignore their numbers):09,10,11 versus OK,OK,OK
Results of weighing 2:

LH: We now know that one of 09,10,11 is heavy. So we use that to eliminate them in another weighing of two of them:Weighing 3: 09 versus 10
Since we’ve already proven the bad coin is heavy, if one side or the other dips, that’s the bad coin. If it doesn’t, the odd coin out (11) is the ‘winner’, and heavy.

RH: The opposite of LH, we know the problem coin is lighter. Do the same test, but look for the scale to tip up to find our coin, or if balanced for it to be coin 11.

EQ: 09,10,11 are equal, so we know it’s 12, but we don’t know if it’s heavier or lighter. For that, just weigh against a good coin and see if it dips up or down:Weighing 3: 12 versus OK coin

With the result of Weighing 1, for both Left Heavier (LH) or Right Heavier (RH), we actually found two items of information from these weighings:

Since one of these eight is the baddy, we’ve proven that the four we left off are ‘good’ coins.

As well, because of this test, if we can somehow know the coin’s number, we’ll also know if it is heavier or lighter – for example, if the left side was heavier, and the coin is 03, we know it must be heavier; if coin 07, lighter.

We now use this info to our advantage. The trick is to swap one coin on either side, and replace the remaining three coins on the right with three of those known good coins (keeping track of every coin of course). So we end up with:

Weighing 2: 01,02,03,05 versus 04,OK,OK,OK
The various possible results are as follows, based on the outcomes from Weighings 1 and 2:

Weighing 1 (LH) Weighing 2 (EQ): if this weighing is equal, the offending coin must be one of the ones we DIDN’T weigh this time – 06,07,08 – AND we know it’s light (since the scale was tipped up on that side in the first weighing). To pick the bad coin from these three, we do our third weighing of two of them:Weighing 3: 06 versus 07
If equal, it’s 08 (and we already know it’s lighter), or if unequal, the lighter one is our culprit.

Weighing 1 (RH) Weighing 2 (EQ): Like the previous case, it’s one of 06,07,08 but this time we know it’s a heavy coin. Weigh 06 versus 07 and pick the heavier one, or if equal, 08.

Weighing 1 (LH) Weighing 2 (LH): We swapped 04 and 05, so if one of them was the bad coin, the scale should have tipped the other direction. It didn’t, so it must be one of 01,02,03 and we know the bad coin must be heavier. We already learned how to weigh and pick a coin from among three when we know if it is heavier or lighter (weigh two, and check for LH/RH/EQ) So we’re done here in three weighings.

Weighing 1 (LH) Weighing 2 (RH): In this case, the scale is now tilting the other way. Now we know it has to be one of the ones we swapped, 04 or 05. So we weigh one against a known good coin:weighing 3: 04 versus OK
If they are equal, then 05 is the bad coin, and from weighing #1, we know it’s light; if not equal, then it’s 04, and heavy.

Weighing 1 (RH) Weighing 2 (LH): Since the scale switched directions between weighings, it’s 04 or 05. Again, doweighing 3: 04 versus OK
and the result is 05+heavy if equal, or 04+light if not (based on Weighing #1).

Weighing 1 (RH) Weighing 2 (RH): Again, swapping 04 and 05 made no difference, so we know it’s 01,02,03 and that one of them is too light (based on weighing #1). Do our 2-coin test for weighing 3, and we’re done.

Whew!

But wait, there’s more:

Problem Three: The Brooklyn Nine-Nine Puzzle

Probably they just renamed the puzzle on the show from coins to men for a few laughs (one character spent time daydreaming who the men would be), but the fact is, that puzzle is not coins in a scale, but men on a seesaw. That makes the puzzle different, not just since you won’t likely get a group of men to balance exactly on a seesaw for weighing purposes (ignoring the practical impossibility of finding 11 men with the same weight to the gram), but because a seesaw is a long beam with many places to sit/weigh from – near the center as well as the far ends.

Of course, if you can get four men to somehow balance properly on a small spot you can solve this puzzle like the previous one, but you might also be able to improve it:

Just draw six lines on each side, evenly spaced from the ends to (almost) the center. All twelve men sit on the seesaw, one on each line. If the seesaw is only slightly unbalanced (the ends don’t go all the way to the ground when everyone sits still), then you have a crude weight measuring device. Have each man swap with an adjacent neighbor, and watch for the change in weight. Only the odd man out swapping will change the scale, and the change will indicate if he’s heavier or lighter than the rest.

One other point: Since the rule is “you only get to use the seesaw three times” and you actually keep everyone on the seesaw the whole time, you’ve actually solved the puzzle in just one ‘use’!

(Addendum – despite the list below, it turns out that eBay DOES have some Brooklyn Nine-Nine
stuff to buy…)

19 thoughts on “Brooklyn 99 And The 12 Men On An Island Puzzle”

your logic here is flawed!
you wrote: “So weigh 6 on one side, 6 on the other. Pick the lighter/heavier side which will contain your ‘bad’ coin.”
you don’t get to “pick” because you don’t know if the scale is heavy on the left OR light on the right…ALL 12 are still in question!!

In problem one you DO know whether the coin is lighter or heavier (“…pretty easy if you know the coin is definitely heavier or lighter than the others…”) – it’s in Problem two that you don’t – as you can see, it makes for a much simpler solution if you know if the extra coin is lighter/heavier in advance…

the first solution i had for the 99 puzzle was another see saw exploit:
Initial state
1-6 against 7-12, but islanders leave the see saw in pairs, ie 1 and 12 get off at the same time 2 and 11 after, when your bad penny exits, you’ll know its one of that pair as the remaining islanders will balance
Theyre the ‘bad pair from here on’

2nd use
you put the bad pair on right against a confirmed pair on left, which gives you the light/heavy bit, if the bad pair are down the bad penny is heavy, up means light

3rd use
Right side bad 1, left bad 2, whichever is ends in the same state as the right in step 2 is your man

(or a confirmed islander works too, since you already know from step 2 whether theyre light or heavy, you just need to find which is different, so if the see saw balances its got to be the unweighed member of the bad pair)

The second is simpler
You’re on an island. Kill them all, and put them in the ocean one by one using the same medium size rock to weight the body each time, then let buouncy figure it out.

Yo Dude, I took another approach and I can’t find a fault in it. I am hoping you might double check it. As it would be crazy if there are 2 correct answers to this puzzle it defies how difficult it is and my assumption is I didn’t take something into account.
After the 1st weigh of 4 vs 4: 1, 2, 3, 4 vs 5, 6, 7, 8 and 9, 10, 11, 12 off to the side.
if it is = then I have the same process of weighing 9, 10, 11, 12.
If it is unequal, left side heavy then (left side 1, 2, 3, 4 is heavy, right side 5, 6, 7, 8 is light).
For move #2: I keep 1, 2 on left, put 3, 4 on the right, move 5 to the left , and keep 6 on the right. 1, 2, 5 vs 3, 4, 6
If it is = then 7 or 8 is the lighter one and simply weigh them against each other for the 3rd move.
A. If left side is heavy then 1 or 2 is the heavy one or 6 is light
A. For move #3: Balance 1 vs 2. If = then 6 is light, If not equal then the heavy one is what we are looking for.
B. If right side is heavy then 3 or 4 is the heavy one or 5 is light.
B. For move #3: Balance 3 vs 4. If = then 5 is light, If not equal then the heavy one is what we are looking for.

The riddle says: “There are 12 men on an island. Eleven weigh exactly the same amount, but one of them is slightly lighter or heavier. You must figure out which. The island has no scales but there is a see-saw. The exciting catch? You can only use it three times.”

The answer: The 12th man is the one who weighs slightly more or slightly less; all that needs to be done is you put 6 men on one side and 5 on the other in addition to the man who weighs slightly more or less (the 12th man), making it six men on each side. If the side with the 12th man on it goes higher, he weighs slightly more, if it goes lower, he weighs slightly less.

The riddle is a simple trick question, much like the following: A British Airways plane flys from London to Egypt. The plane crash lands in Italy. Assuming all the passengers are either Egyptian or British, what country’s responsibility is it to bury the survivors?

Yes, that would work if you know the man is definitely lighter or definitely heavier. A problem exists if you don’t know heavier/lighter – for example, did the left side dip because the (heavier) man was on it, or because the (lighter) man was on the right? With only the information from the weighing (that is, we’re not eyeing them to see who is bigger/smaller) we need more than a single weighing.

P.S. For anyone reading/wondering, the trick mentioned for the flight in the last comment was in the word ‘survivors’ – you don’t bury survivors (at least, hopefully not for many, many years!)

That’s the trick of this puzzle – trying to get extra information in minimal guesses. Think of it this way: If you had three people, you can get the information in three weighing or less: A to B, B to C, and A to C (actually, 2 would do, but explaining how gets as complicated as the puzzle above). One of those weighings would be equal, but two would be off. But the equal weighing helps, since it would tell us the ‘right’ weight, and gives us information for the other weighings. For example, if A=B, then if C tips low against A, it’s heavier; if it tips high, it’s lighter. The puzzle just expands on that by trying to minimize the number of weighings.

Thinking outside the box. Looked up the definition of seesaw. Don’t let the atypical pic set the precedence. It’s the definition that matters. If the seesaw was actually a round disc balanced on a cone-like fulcrum, then the solution is simple. Place all 12 men on the perimeter of the disc equidistant apart. Theoretically, the disc will tip in a “seesaw” manner. At the point where the disc touches the ground, the individual on that point is either heavier than the other 11 or the individual at the opposite end of that point is lighter than the other 11. Leave those two on the disc and remove the other 10. They weigh the same. Now, is the 12th man heavier or lighter? Remove one of the two individuals on the seesaw and replace him with one of the 10 previously removed. If the seesaw doesn’t change, then the man at the other end is lighter than the 11. If the seesaw balances itself, then the man removed was heavier. Solved in two moves….